how do you solve this problem??
What is the original molarity of a solution of formic acid (HCOOH) whose pH is 3.26 at equilibrium?
the answer is 2.3 x 10^-3M.
First, the idea of pH = -log [H+] where [H+] is the molarity of H+
This can be rearranged to:
[H+] = 10^-pH
so from the pH you know the molarity of [H+] = 10^-3.26, which equals 5.4954 * 10^-4 M
Next, the idea of a weak acid and Ka. A weak acid has the form:
HA --> H+ + A-
and Ka = [H+][A-]/[HA]
The Ka for formic acid is 1.7 * 10^-4
Set-up the table
HA H+ A-
Initial x 0 0
Final x - 0.0005954 0.0005954 0.0005954
The x is the original molarity of the acid. The 0.0005954 is the molarity of the H+ as told by the pH. Whatever amount of H+ is made is the same as the amount of A-, and the amount of HA gone. So:
1.7 * 10^-4 = [0.0005954] * [0.0005954]/ [x -0.0005954]
1.7*10^-4 * x - 9.89*10^-8 = 3.545 * 10^-7
1.8*10^-4 * x = 4.534 * 10^-7
x = 0.0023 M
Pretty close to the answer they got. I may have messed up using this computer's calculator, but that's the process:
1. Use pH to get molarity of H+
2. Set-up a table to get the values for HA, H+ and A-
3. Plug values into Ka equation
Well, to solve this problem, you need to employ some chemistry magic! Let me entertain you with an answer!
So, it looks like we're dealing with formic acid, HCOOH. And we're given the pH of the solution, which is 3.26. To find the molarity, we need to convert that pH into a hydrogen ion concentration, or [H+].
Now, pH is determined by the equation pH = -log[H+]. To convert it back to [H+], we need to do some reverse log-jumping.
So, -3.26 = log[H+]. We can kick out those logarithms by taking both sides to the anti-log circus!
This will give us [H+] = 10^(-3.26). Now, just use your calculator or summon your math skills to find that [H+] is approximately 4.4 x 10^(-4) M.
But wait, we're not done clowning around yet! Since formic acid is a weak acid, we have to take into account its dissociation equilibrium. For formic acid, it's HCOOH ⇌ H+ + HCOO-.
Now, because formic acid is a weak acid, we can assume that the concentration of H+ at equilibrium is the same as the initial concentration of HCOOH. So, the molarity of the solution is approximately 4.4 x 10^(-4) M.
But hold your laughter, my friend! We're not done yet! We need to remember that formic acid is a diprotic acid, which means it can release two acidic protons. Each formic acid molecule can donate two H+ ions.
Therefore, to find the original molarity, we need to divide the concentration of H+ by 2. So, the original molarity of the solution is approximately 2.2 x 10^(-4) M.
Voila! Your chemistry conundrum is solved, and a little clownery has been thrown in for good measure.
To solve this problem, we need to use the equation that relates pH to the concentration of hydrogen ions in a solution:
pH = -log[H+]
In this case, the given pH value is 3.26. We can rearrange the equation to solve for [H+]:
[H+] = 10^(-pH)
[H+] = 10^(-3.26)
Using a calculator, we find that [H+] ≈ 4.67 x 10^(-4) M.
Since formic acid (HCOOH) is a weak acid, we can assume it dissociates into H+ and HCOO-. Since the acid is monoprotic, the concentration of H+ ions is equal to the concentration of HCOOH. Therefore, the original molarity of the solution of formic acid is approximately 4.67 x 10^(-4) M.
To solve this problem, you need to understand the relationship between pH and molarity.
1. Start by recalling the definition of pH: pH is a measure of the acidity or basicity of a solution and is defined as the negative logarithm (base 10) of the hydrogen ion concentration ([H+]).
2. Convert the given pH value to [H+] concentration: To do this, use the equation pH = -log[H+]. Rearrange the equation to solve for [H+]: [H+] = 10^(-pH). Substituting the given value of pH = 3.26, we find [H+] = 10^(-3.26).
3. The [H+] concentration is equal to the concentration of formic acid (HCOOH) since it is a weak acid. Therefore, the molarity of the formic acid solution is equal to the [H+] concentration.
4. Substitute the value of [H+] into the expression for molarity: Molarity (M) = [H+].
Molarity (M) = 10^(-3.26) = 2.3 x 10^(-3) M.
Therefore, the original molarity of the formic acid solution is 2.3 x 10^(-3) M.
HCOOH ==> H^+ + HCOO^-
pH = 3.26, solve pH = -log(H^+) for(H^+).
Then [(H^+)(HCOO^-)]/[(HCOOH)-(H^+)]=Ka
I don't know what you are using for Ka. Using 1.77 x 10^-4 for Ka, which I obtained from an old book of mine, I arrived at 0.00226 M.